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Helllo, I'm new here. I will try this great place to get my answer. I have a next problem:

$$\vartheta = \arctan(K \tan \varphi)$$

$K > 0$

Where $K$ is a constant. If $K = 1$ then $\vartheta = \varphi$, but I can't find more simple answer for $\vartheta$ if $K \neq 1$. Maybe someone will be able to help me out, if it is even possbile.

Michael Hardy
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Grega
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    A solution to $\vartheta = \arctan(K \tan \varphi)$ is an intersection of the graphs of $\tan x$ and $K\tan x$ in the interval $(-\pi/2,\pi/2)$. But there is only one such intersection when $K \neq 1$: the intersection at $x = 0$. See this plot. – Antonio Vargas Feb 23 '12 at 16:06
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    So what is your question? What do you need to derive? Do you have to find $\theta$ ? – Kirthi Raman Feb 23 '12 at 16:00
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    @Antonio: The question asks for $\vartheta=\arctan(K\tan(\varphi))$ not $\varphi=\arctan(K\tan(\varphi))$. – robjohn Feb 23 '12 at 16:13
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    @AntonioVargas it looks like you're finding values of x for which $\tan(x)$ and $K\tan(x)$ are equal, but the OP is seeking the relationship between $x$ and $y$ where $\tan(x)=K\tan(y)$. For example, $\theta = \pi/3$,$\phi = \pi/6$, $K=3$ works. – Aru Ray Feb 23 '12 at 16:15
  • Thanks for answers. I would like to get simpler form for $\vartheta$ if $K \neq 1$. I was wondering if there exists any derivation of that form. – Grega Feb 23 '12 at 16:16
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    There are formulas for $\tan(K\arctan(x))$ for certain values of $K$, but I don't know of any for $\arctan(K\tan(\varphi))$. – robjohn Feb 23 '12 at 16:17
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    @robjohn: Whoops, you're right. – Antonio Vargas Feb 23 '12 at 16:17
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    http://www.wolframalpha.com/input/?i=%5Ctheta+%3D+arctan%28K+tan%28%5Cphi%29%29 – Kirthi Raman Feb 23 '12 at 16:25
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    Grega, what do you mean by "simpler"? We could, for example, use the exponential form of $\tan$ to derive the logarithmic form of $\arctan$, then combine these in $\vartheta = \arctan(K \tan \varphi)$ and simplify to get $$\vartheta = \frac{i}{2} \text{Log}\left(\frac{1-e^{2 i \varphi} (K-1)+K}{1+e^{2 i \varphi} (K+1)-K}\right).$$ – Antonio Vargas Feb 23 '12 at 16:44
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    If $K$ is a positive integer, then $\tan(Kx)$ can be expressed as a not horrible rational function of $\tan x$. You then end up solving a polynomial equation of degree $K$. I doubt there is a general expression for the roots. For non-integer $K$, the problem is likely to be hopeless. – André Nicolas Feb 23 '12 at 16:49
  • Ok. Thanks for answers. I knew if there was anything simple possible I would find it already. – Grega Feb 23 '12 at 16:51
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    If an approximation is enough,$$\vartheta=K,\varphi+\frac{1}{3} \left(K-K^3\right)\varphi^3 + O(\varphi^5)$$ – Julián Aguirre Feb 23 '12 at 17:31
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    If the tan and arctan were reversed, it would be a lot easier. I know of no way offhand to represent $\arctan kx$ in terms of $\arctan x$ – Mike Feb 23 '12 at 19:36
  • Thanks for all the answers. I knew I have a tough one. I just left it the way it is, because in that way it bothers my derivations at a minimum :). – Grega Feb 24 '12 at 22:29
  • Since wolframalpha is providing exact solutions, there must be a better method... – Gregory Morse May 24 '23 at 23:06

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